Timeline for A map from a symmetric product of a curve to its Jacobian
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Jun 11, 2019 at 15:35 | comment | added | abx | Sorry, but if you don't see why $H_1$ is relevant, there is no way I can explain it in a comment. Have a look at the chapter on Riemann surfaces in Griffiths and Harris. | |
| Jun 11, 2019 at 14:43 | comment | added | Alessio | @abx Thanks for the answer. Sorry, but I don't get your hint, can you spell out a bit more? (Meanwhile I figured out another proof by showing that $J$ satisfies the universal property of $\mathrm{Alb}(C^{(n)})$ but I still would like to understand your approach). I tried to use that $\mathrm{Alb}(X)=\mathrm{Pic}^0(X)^{\vee}$ and the exponential sequence, but I don't see where the isomorphism on $H_1(-,\mathbb{Z})$ comes from and how to apply it. | |
| Jun 10, 2019 at 16:44 | comment | added | abx | @Alessio: Yes, $\operatorname{Alb}(C^{(n)})\cong J $. Use the fact that the natural map $C^{(n)}\rightarrow J$ induces an isomorphism on $H_1(-,\mathbb{Z})$. | |
| Jun 10, 2019 at 9:49 | comment | added | Alessio | @abx Sorry for the intrusion, but why is $\mathrm{Alb}(C^{(n)}\times C^{(n)})\cong J\times J$? Is in general, $\mathrm{Alb}(C^{(n)})\cong J$? Where does it come from? | |
| Feb 4, 2018 at 17:01 | comment | added | abx | Sorry, I assumed the OP meant projective fiber bundle. I am not sure what he means by "fiber bundle". | |
| Feb 4, 2018 at 16:31 | answer | added | Sasha | timeline score: 5 | |
| Feb 4, 2018 at 16:25 | comment | added | Jason Starr | I also agree that this is not a "fiber bundle" in the sense of a morphism that, locally on the target, is isomorphic to projection of a product scheme to a factor. However, I do not see how to deduce this from the Albanese variety (perhaps the fiber itself has nontrivial Albanese isomorphic to $J$). Rather, I believe that this follows from the fact that the projective bundle $C^{(n)}\to \text{Pic}^n_{C/k}$ is not equivariant for the natural action of $\text{Pic}^0_{C/k}$. Of course if $k$ equals $\mathbb{C}$, then Ehresmann applies to the underlying differentiable manifolds. | |
| Feb 4, 2018 at 13:56 | comment | added | abx | Certainly not. If it was, the induced map on the Albanese variety would be an isomorphism. But $\operatorname{Alb}(C^{(n)}\times C^{(n)})\cong J\times J $. | |
| Feb 4, 2018 at 8:12 | history | asked | user4231 | CC BY-SA 3.0 |